Quantities and Concentrations

SI Base UnitsPhysical quantity Name of Units Abbreviation

Mass kilogram kg

Length meter m

Time second s

Temperature kelvin K

Amount of substance mole mol

Electric current ampere A

Luminous intensity candela cd

SI Prefixes

especially useful in this course giga G 109

mega M 106

kilo k 103

centi c 10-2

milli m 10-3

micro 10-6

nano n 10-9

pico p 10-12

Quantities

• Mole

• Molar mass

• Mass

• Weight

• What is the difference between mass and weight?

Useful Algebraic Relationships

wt A (g) mol A = -----------------

fw A (g/mol)

mol A = V (L) x M (mol A/L soln)

or

wt A (mg) mmol A = -----------------

fw A (g/mol)

mmol A = V (mL) x M (mmol A/mL soln)

Solution Terminology

• solute

• solvent

• aqueous solution

• liter

• atomic weight

• molecular weight

Molar concentrtion-Molarity

no. moles AMolarity => M = -------------------

no. liters solutionor

no. millimoles AMolarity => M = -------------------------

no. milliliters solution

Analytical Molarity Equilibrium molarity

• What are analytical molarity and equilibrium molarity?

• What is the difference between them?

Percent Composition

wt of a solutew - w% = -------------------- × 100%

wt of solution

vol of a solutev - v% = -------------------- × 100%

vol of solution

wt of a solutew - v% = --------------------- × 100%

vol of solution

p-Functions

pX = - log10[X]

examples:

pH

pOH

pCl

pAg

Parts per Million / Billion

wt of a solutecppm = ------------------- × 106

wt of solution

wt of a solutecppb = ------------------- × 109

wt of solution

Empirical FormulasMolecular FormulasStructure Formulas

Chemical Stoichiometry

Preparing Solutions

EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.

EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.

(1.00 L soln) # g NaOH = ----------------

EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.

(1.00 L soln)(0.100 mol NaOH)

# g NaOH = --------------------------------------- (1 L soln)

EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.

(1.00 L soln)(0.100 mol NaOH)

# g NaOH = --------------------------------------- (1 L soln)

EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.

(1.00)(0.100 mol NaOH)# g NaOH = ------------------------------

(1)

EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.

(1.00)(0.100 mol)(40.00g NaOH)

# g NaOH = ---------------------------------------- (1) (1 mol)

EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.

(1.00)(0.100 mol)(40.00g NaOH)

# g NaOH = ---------------------------------------- (1) (1 mol)

EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.

(1.00)(0.100)(40.00g NaOH)# g NaOH = ----------------------------------------

(1) (1)

EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.

(1.00)(0.100)(40.00g NaOH)# g NaOH = ----------------------------------------

(1) (1)

= 4.00 g NaOH

EXAMPLE: Describe the preparation of 1.00 L of 0.100 M NaOH solution (f.w. 40.00) from reagent grade solid.

(1.00)(0.100)(40.00g NaOH)# g NaOH = ----------------------------------------

(1) (1)

= 4.00 g NaOH

Weigh 4.00 g of NaOH, transfer to a 1.00 L volumetric flask, and dilute to the line.

Dilution

#moles solute in conc. soln

equals

#moles solut in dil. soln

therefore

Mconc Vconc = Mdil Vdil